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Nuclear Decay

  1. Sep 15, 2008 #1
    1. The problem statement, all variables and given/known data

    In a sample of one litre of carbon dioxide at STP an average of 5 disintegrations

    C -> N + e + v

    are observed per minute. Calculate the atomic fraction of C present if the mean lifetime of this nucleus is 8267 years.

    2. Relevant equations

    [tex]\tau[/tex] = [tex]\omega[/tex]-1 = 8267 years

    N(t) = N0 * [tex]e[/tex]-[tex](\omega[/tex] * t)

    3. The attempt at a solution

    First I found [tex]\tau[/tex] = 2.607 * 1011 seconds.

    And really I'm at a loss of where to go next. I figure N(t)/N0 is what im looking for but I cannot solve for this without knowing the time t and I do not know where the 5 disintegrations per minute comes in. I'm not looking for an answer(already know it), i just want a little push in the right direction.
  2. jcsd
  3. Sep 15, 2008 #2


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    Homework Helper
    Gold Member

    Since there are an average of 5 disintegrations per minute, you would expect:

    [tex] N(t)-N(t+60s)=5 [/tex]...can you take it from there?
  4. Sep 15, 2008 #3


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    Staff: Mentor

    One can find the total amount of carbon from "one litre of carbon dioxide" - simply find the moles of CO2 in 1 liter at STP. It would appear that one is to find the atomic fraction of radiocarbon or C-14.

    The activity, decays per unit time, is the product of the decay constant [itex]\lambda[/itex] and the number of atoms of the radionuclide present, i.e. A(t) = [itex]\lambda[/itex] N(t). One is given A, so find N.
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